Question

Pure $O_2$ diffuses though an aperture in $224$ sec, whereas mixture of $O_2$ and another gas containing $80$% $O_2$ takes $234$ sec to effuse out same volume. What is molecular weight of the gas?

• A. 61.5 g
• B. 51.5 g
• C. 41.5 g
• D. 31.5 g

Answer 1

Answer: (C)

41.5 g

For gaseous mixture $80\%O_2,and20\%$ of other gas

Let molecular weight of gas be $m$

Average molecular weight $M_m=32\left(for{O}_2\right)\times 0.80+0.20\times m$

Now, for diffusion of gaseous mixture and pure $O_2$

$\frac{r_{O_2}}{r_m}=\sqrt{\frac{M_m}{M_{O_2}}}$

$\frac{1/224}{1/234}=\sqrt{\frac{140}{28}}$

$M_m=33.9$

$32\times 0.80+0.20\times m=33.9$

$m=\frac{33.9-25.6}{0.20}$

$m=41.5g$